Steps of Construction:
i) Draw a circle with radius = 4 cm.
ii) Take a point A on it.
iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.
iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
v) Join AB and AC.
vi) Draw the perpendicular bisector of AC, which intersects AC at M and meets the circle at E and F.
EF is the locus of points inside the circle which are equidistant from A and C.
vii) Join AE, AF, CE and CF.
Proof :
i) In ΔCME and ΔAME
CM = AM (EF is the bisector of AC)
∠CME = ∠CMA = 90°
EM = EM (Common)
∴ By side Angle side criterion of congruence,
ΔCME ≅ ΔAME ( SAS Postulate)
The corresponding parts of the congruent triangles are congruent.
⟹ CE = AE (CPCT)
Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.
Therefore, Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.
